ГДЗ по алгебре 9 класс Макарычев Задание 810

 

\[\boxed{\text{810\ (810).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[0 < P(A) < 1;\]

\[0 < P(B) < 1;\]

\[0 < P(C) < 1;\]

\[P(D) = 0.\]

\[\boxed{\text{810.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ (2x + 1)(x + 4) - 3x(x + 2) < 0\]

\[2x^{2} + 8x + x + 4 - 3x^{2} - 6x < 0\]

\[x^{2} - 3x - 4 > 0\]

\[(x + 1)(x - 4) > 0\]

\[x \in ( - \infty; - 1) \cup (4;\ + \infty).\]

\[\textbf{б)}\ (3x - 2)^{2} - 4x(2x - 3) > 0\]

\[9x^{2} - 12x + 4 - 8x^{2} + 12x > 0\]

\[x^{2} > - 4\]

\[x \in ( - \infty; + \infty).\]

\[1 - 36x^{2} + 35x^{2} - 14x > 14\]

\[x^{2} + 14x + 13 < 0\]

\[(x + 1)(x + 13) < 0\]

\[x \in ( - 13; - 1).\]

\[5x² - 5x + 2x - 2 - 4x^{2} + 1 < 27\]

\[x^{2} - 3x - 28 < 0\]

\[(x - 7)(x + 4) < 0\]

\[x \in ( - 4;7).\]

\[\textbf{д)}\ (2x - 1)(1 + 2x) - x(x + 4) < 6\]

\[2x + 4x^{2} - 1 - 2x - x^{2} - 4x - 6 < 0\]

\[3x^{2} - 4x - 7 < 0\]

\[(x + 1)\left( x - 2\frac{1}{3} \right) < 0\]

\[- 1 < x < 2\frac{1}{3}.\]

\[\textbf{е)}\ (3x - 1)x - (6 - x)(x + 6) < 37\]

\[3x^{2} - x - 36 + x^{2} - 37 < 0\]

\[4x^{2} - x - 73 < 0\]

\[x \in \left( \frac{1 - \sqrt{1669}}{8};\ \frac{1 + \sqrt{1669}}{8} \right).\]